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3 years ago

Which of these solutions has the highest freezing point?

2 answers:

8 0

The right answer for the question that is being asked and shown above is that: "1.0 M ionic aluminum bromide (AlBr3)." the solution that has the highest freezing point is that 1.0 M ionic aluminum bromide (AlBr3)

ANEK [815]3 years ago

5 0

Answer: 1.0 M molecular sucrose (C₁₂H₂₂O₁₁).

Explanation:

1) The depression of the freezing point of a solvent when you add a solute is a colligative property.

2) Colligative properties are those physical properties of solutions that depends on the number of solute particles dissolved into the solution.

3) The relation between the number of solute particles and the depresson of the freezing point is proportional: the greater the number of solute particles the greater the freezing point depression.

4) You need to find the solution with the highest freezing point, this is the solution in which the freezing point decreased the least.

5) Then, that is the solution with least number of solute particles.

6) Since all the given solutions have the same molarity (1.0 M), you only have to deal with the possible ionization of the different solutes.

7) NaCl, CaBr₂, AlBr₃, and KCl are ionic compounds, so each unit of them will ionize into two, three, four, and two ions, respectively, while sucrose, being a covalent compound does not dissociate.

Then, 1.0 M solution of sucrose will have less solute particles than the others, and will exhitibit the lowest freezing point depression, meaning that it will have the highest freezing point of the given solutions.

You might be interested in

What is the chemical formula of the salt produced by the neutralization of potassium hydroxide with sulfuric acid?KSO3KSO4K(SO4)

kakasveta [241]

Answer: The salt produced will be $K_{2}SO_{4}$

Explanation:

During a neutralization reaction, an acid reacts with a base for producing the correspondent salt, and water.

The strong acids release all the protons avalaible when are dissolved, such as sulfuric acid. As you can see, sulfuric acid have 2 protons ready for being released ( $H_{2}SO_{4}$ ); and those places have to be occcupied for other ions equivalents to the H+: K+ from KOH in this case.

Therefore the answer will be $K_{2}SO_{4}$ .

6 0

3 years ago

What do you call the procedure that helps you determine the volume of an irregularly shaped object, whole using a graduated cyli

ki77a [65]

The method is called the displacement method.

You place some water in the graduated cylinder and measure its volume.
Then you add your object and measure the new volume.
The difference between the two volumes is the volume of your object.

6 0

3 years ago

A buffer solution contains 0.479 M NaHCO3 and 0.342 M Na2CO3. Determine the pH change when 0.091 mol HNO3 is added to 1.00 L of

pshichka [43]

Answer:

ΔpH = 0.20

Explanation:

The buffer of HCO₃⁻ + CO₃²⁻ has a pka of 10.2

HCO₃⁻ ⇄ H⁺ + CO₃²⁻

There are 0.479moles of NaHCO₃ and 0.342moles of Na₂CO₃.

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.342mol / 0.479mol

pH = 10.05

NaOH reacts with HCO₃⁻ producing CO₃²⁻, thus:

NaOH + HCO₃⁻ → CO₃²⁻ + H₂O + Na⁺

0.091 moles of NaOH produce the same moles of CO₃²⁻ and consume HCO₃⁻. Moles of these species are:

CO₃²⁻: 0.342mol + 0.091mol: 0.433mol

HCO₃⁻: 0.479mol - 0.091 mol: 0.388mol

Using Henderson-Hasselbalch formula:

pH = pka + log [Base] / [Acid]

pH = 10.2 + log 0.433mol / 0.388mol

pH = 10.25

That means change of pH, ΔpH is:

ΔpH = 10.25 - 10.05 = 0.20

I hope it helps!

3 0

3 years ago

Answer the following questions:

Vera_Pavlovna [14]

Answer:

fdfdgiukjgrguhjnfvthhhdqddwsvhjjj

4 0

2 years ago

Read 2 more answers

A 10.21 mol sample of argon gas is maintained in a 0.7564 L container at 296.9 K. What is the pressure in atm calculated using t

soldi70 [24.7K]

Answer:

The pressure in atm calculated using the van der Waals' equation, is 337.2atm

Explanation:

This is the Van der Waals equation for real gases:

(P + a/v² ) ( v-b) = R .T

where P is pressure

v is Volume/mol

R is the gas constant and T, T° in K

a y b are constant for each gas, so those values are data, from the statement.

[P + 1.345 L²atm/mol² / (0.7564L/10.21mol)² ] (0.7564L/10.21mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

[P + 1.345 L²atm/mol² / 5.48X10⁻³ L²/mol²] (0.074 L/mol - 3.219×10-2 L/mol ) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 0.082 L.atm/mol.K . 296.9K

(P + 245.05 atm) (0.04181L/mol) = 24.34 L.atm/mol

0.04181L/mol .P + 10.24 L.atm/mol = 24.34 L.atm/mol

0.04181L/mol .P = 24.34 L.atm/mol - 10.24 L.atm/mol

0.04181L/mol. P = 14.1 L.atm/mol

P = 14.1 L.atm/mol / 0.04181 mol/L

P = 337.2 atm

4 0

3 years ago